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Question

Question: How do you solve the quadratic equation \[100{{x}^{2}}-800x+1500=0\]?...

How do you solve the quadratic equation 100x2800x+1500=0100{{x}^{2}}-800x+1500=0?

Explanation

Solution

This question is from the topic of algebra. In this question, we will find the value of x. In solving this question, we will first take out the common term 100 and remove the term 100 from the equation. After that, we will solve the quadratic equation. We will use here Sridharacharya’s Rule to solve the quadratic equation. After solving the quadratic equation, we will get our answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to solve the equation 100x2800x+1500=0100{{x}^{2}}-800x+1500=0.
So, the equation which we have to solve is
100x2800x+1500=0100{{x}^{2}}-800x+1500=0
We can write the above equation as
100x2100×8x+100×15=0\Rightarrow 100{{x}^{2}}-100\times 8x+100\times 15=0
Now, we will take out the common term 100 from the above equation. We can write the above equation as
100(x28x+15)=0\Rightarrow 100\left( {{x}^{2}}-8x+15 \right)=0
Now, we will divide the term 100 to the both side of equation, we can write
(x28x+15)=0100\Rightarrow \left( {{x}^{2}}-8x+15 \right)=\dfrac{0}{100}
As we know that the 0 divided by any real number is zero, so we can write
x28x+15=0\Rightarrow {{x}^{2}}-8x+15=0
Now, we will find the roots or we can say that we will find the values of x from the above equation using Sridharacharya’s Rule.
So, according to the Sridharacharya’s Rule, we can write the value of x as
x=(8)±(8)24×1×152×1x=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 1\times 15}}{2\times 1}
The above can also be written as
x=8±64602\Rightarrow x=\dfrac{8\pm \sqrt{64-60}}{2}
The above can also be written as
x=8±42\Rightarrow x=\dfrac{8\pm \sqrt{4}}{2}
The above can also be written as
x=8±22\Rightarrow x=\dfrac{8\pm 2}{2}
The above can also be written as
x=4±1\Rightarrow x=4\pm 1
From the above, the values of x are 5 and 3.
Now, we have solved the equation 100x2800x+1500=0100{{x}^{2}}-800x+1500=0 and got the values of x as 5 and 3.

Note: We should have a better knowledge in the topic of quadratic equation which belongs to the chapter algebra. We should know how to find the roots of the quadratic equation using different methods. We should remember the Sridharacharya’s Rule and the terms correctly. We can also solve without taking 100 outside, but then we will end up with bigger values of a, b, c and it will make calculations complicated. There is another method - factorisation which we can apply to get the roots.